\((0, \ y)\) is some point on the y-axis. In the drawing above, this is shown in the smaller washer off to the side as the distance between the points labeled \((0, \ y)\) and \((x, \ y)\). The inner radius of a washer will be the distance between the center of the washer and the inner edge. This will be used to help us find the inner and outer radii of the washers. Take a look at the smaller washer in the upper left section of our graph. The large washer in the middle of our graph is there to help you visualize where these washers would be if we were to stack them up to create this figure. Remember, as we showed in the first washer method practice problem, the volume of a washer is given by \(V=\pi h(R^2-r^2)\) where r is the inner radius and R is the outer radius. Let’s take a minute to consider the dimensions of this particular washer. You can see one of these infinitely thin washers drawn in the figure. How do we set the integral up with respect to y? Therefore, when we create our integral, it will all need to be in terms of y rather than x. As a result of this, we will be integrating with respect to y ! Since we move in the y direction to get to the next washer, we need to integrate with respect to y. This is an important difference because adding up the volume of all of these washers will require us to move vertically throughout this figure to get the next washer and add its volume to the total. This is different from the first washer method example I did, where the washers were all side by side. You will notice that if we imagine this figure as a stack of washers, the washers would be stacked vertically, one on top of the other. What we need to think about is a stack of very, very, very thin washers stacked one on top of the other, in the same shape as the figure shown a couple paragraphs ago. This is important to distinguish here because we need to imagine all of the washers that make up this 3-D figure. For this example, we will proceed using the washer method.
We could solve this problem using the cylinder method as well, but that’s for another lesson. All of the graphing and sketching is to help us visualize what is being described so we can correctly formulate our integral. This step is at the heart of these problems. Result of rotating the region about the y-axis.
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